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### Transcript

• 1. Physics Applied to Radiology Chapter 3 Fundamentals of Physics
• 2. Physics
• natural science
• deals with matter and energy
• defines & characterizes
• interactions between matter and energy
• 3. Matter
• a physical substance
• characteristics of all matter
• occupies space
• has mass
• 4. Energy
• capacity for doing work
• 5. Math
• exact vs. approximate numbers
• exact -- defined or counted
• approximate -- measured
• examples
• # of chairs in room
• # of seconds in a minute
• # seconds to run 100 m dash
• 6.
• # of digits in a value when...
• leading & trailing zeros are ignored
• trailing 0 may be designated as significant
• the decimal place is disregarded
• How many significant figures?
• Value: significant figures
• 3.47
• 0.039
• 206.1
• 5.90
Significant Figures
• 7.
• # of digits in a value when...
• leading & trailing zeros are ignored
• trailing 0 may be designated as significant
• the decimal place is disregarded
• How many significant figures?
• Value: significant figures
• 3.47 3
• 0.039 2
• 206.1 4
• 5.90 2
Significant Figures
• 8. Accuracy vs. Precision
• accuracy -- # of significant figures
• 3.47 is more accurate than 0.039
• precision -- decimal position of the last significant figure
• 0.039 is more precise than 3.47
• 9. Example
• Describe the accuracy and precision of the following information.
• 2.5 cm metal sheet with a .025 cm coat of paint
• accuracy is same for both (2 sig. fig.)
• precision is > for paint (1/1000 vs. 1/10)
• 10. Rounded Numbers
• all approximate # are rounded
• last digit of approx. number is rounded
• last sig. fig. of an approx. # is never an accurate #
• error of last number is ½ of the last digit's place value
• (if place value is .1 then error = .05)
• 11. Rounded Number
• example:
• if a measured value = 32.63
• error is .005 (½ of .01)
• actual # is between
• 32.635 (32.63 + .005)
• 32.625 (32.63 - .005)
• 12. Rounding Rules
• round at the end of the total calculation
• do not round after each step in complex calculations
• when - or + use least precise #
• (same # of decimal places)
• when x or ÷ use least accurate #
• (same # of sig. figures)
• 13. Rounding Example 1
• 73.2
• 8.0627
• 93.57
• + 66.296
• 241.1287
• 241.1 # decimal places = to least precise value
• 14. Rounding Example 2
• 2.4832
• x 30.51
• 75.762432
• 75.76 # significant figures = to least accurate number
• 15. Numerical Relationships
• direct linear
• as x  y  (or vice versa)
• example formula y = k x
• expressed as proportion y  x
• example: x y (for y = 5x)
• 1 5
• 2 10
• 3 15
• 16. Numerical Relationships
• direct exponential
• direct square (or other exponent)
• as x  y  by an exponential value  (or vice versa)
• example formula y = k x 2
• expressed as proportion y  x 2
• example: x y (for y = 5x 2 )
• 1 5
• 2 20
• 3 45
• 17. Numerical Relationships (cont.)
• indirect
• as x  y 
• example formula x y = constant
• expressed as proportion y  1/x
• example: x y (for xy = 100)
• 1 100
• 2 50
• 4 25
• 18. Numerical Relationships (cont.)
• indirect exponential
• inverse square (or other exponent)
• as x  y  by an exponential value  (or vice versa)
• example formula y x 2 = constant
• expressed as proportion y  1/ x 2
• example: x y (for x 2 y = 100)
• 1 100
• 2 25
• 4 6.25
• 19. Graphs
• used to display relationships between 2 variables
• Y-axis (dependent) measured value
• X-axis (independent) controlled value
x-axis y-axis
• 20. Graphic Relationships ( on linear graph paper)
• slope (left to right)
• direct = ascending
• indirect = descending
• shape
• linear = straight
• exponential = curved
• 21. Evaluating Graphed Information
• identify variables
• describe shape & slope of line
• correlate information to theory
• 22. Example #1
• Relationship of mA to Intensity
• 23. Example #1 (evaluated)
• Relationship of mA to Intensity
• variables
• independent = mA
• dependent = Exposure
• shape & slope
• slope = ascending (=direct)
• shape = straight line (=linear)
• correlate to theory
• mA has a direct linear relationship to exposure; as mA increases exposure increases in a similar fashion; the graph demonstrates that if you double the mA (200 to 400) you also double the exposure (30 mR to 60 mR )
• 24. Example #2
• Relationship of the # days before exam to amount of study time
• 25. Quantities & Units
• quantity = measurable property
• quantity definition (what is measured)
• length distance between two points
• mass amount of matter (not weight)
• time duration of an event
• unit = standard used to express a measurement
• quantity unit other units
• length meter
• mass kilogram
• time second
• 26. Unit Systems
• System length mass time
• English foot slug (pound) second
• metric SI** meter kilogram second
• ** also ampere, Kelvin, mole, candela
• metric MKS meter kilogram second
• metric CGS centimeter gram second
• Do not mix unit systems when doing calculations!!
• 27. Converting Units
• convert 3825 seconds to hours
• identify conversion factor(s) needed
• factors needed: 60 sec = 1 min & 60 min = 1 hour
• arrange factors in logical progression
• For seconds  hours
• sec  min/sec  hour/min
• set up calculation
• 28. Dimensional Prefixes Bushong, table 2-3 (pg 23)
• used with metric unit systems
• modifiers used with unit
• a power of 10 to express the magnitude
• prefix symbol factor numerical equivalent
• tera- T 10 12 1 000 000 000 000
• giga- G 10 9 1 000 000 000
• mega- M 10 6 1 000 000
• kilo- k 10 3 1 000
• centi- c 10 -2 .01
• milli m 10 -3 .001
• micro-  10 -6 .000 001
• nano- n 10 -9 .000 000 001
• pico- p 10 -12 .000 000 000 001
• 29. Rules for Using Prefixes
• To use a prefix divide by prefix value & include the prefix with the unit
• To remove a prefix multiply by prefix value & delete prefix notation from the unit
• 30. Base Quantities & Units (SI)
• describes a fundamental property of matter
• cannot be broken down further
• quantity SI unit definition for quantity
• length meter distance between two points
• mass kilogram amount of matter (not weight)
• time second duration of an event
• 31. Derived Quantities & Units
• properties which arrived at by combining base quantities
• quantity units definition for quantity
• area m x m m 2 surface measure
• volume m x m x m m 3 capacity
• velocity m/s m/s distance traveled per unit time
• acceleration m/s/s m/s 2 rate of change of velocity
• ms -2
• 32. Derived Quantities with Named Units
• quantities with complex SI units
• quantity units definition
• frequency Hertz Hz # of ?? per second
• force Newton N &quot;push or pull&quot;
• energy Joule J ability to do work
• 33. Solving Problems
• 1. Determine unknown quantity
• 2. Identify known quantities
• 3. Select an equation (fits known & unknown quantities)
• 4. Set up numerical values in equation
• same unit or unit system
• 5. Solve for the unknown
• write answer with magnitude & units
• 34. Mechanics
• study of motion & forces
• motion = change in position or orientation
• types of motion
• translation
• one place to another
• rotation
• around axis of object's mass
• 35. Measuring Quantities in Mechanics
• all have magnitude & unit
• scalar vs. vector quantities
• Scalar -- magnitude & unit
• Vector -- magnitude, unit & direction
run 2 km vs run 2 km east
• requires use of graphs, trigonometry or special mathematical rules to solve
• example:
F 1 F 2 F 1 + F 2 = Net force
• 37. Quantities in Mechanics
• speed
• rate at which an object covers distance
• rate
• indicates a relationship between 2 quantities
• \$/hour exams/tech # of people/sq. mile
• speed = distance/time
• speed is a scalar quantity
• 38. Speed (cont.) d in m t in s v = m/s same at all times total distance total time General Formula: Variations: instantaneous uniform average v at 1 point in time v = d t distance time
• 39. Speed Example
• An e - travels the 6.0 cm distance between the anode & the cathode in .25 ns. What is the e - speed? [Assume 0 in 6.0 is significant]
• v = ?? 6.0 cm = distance .25 ns = time
• v = d / t (units: m /s  need to convert)
• 6.0 cm = 6.0 x 10 -2 m .25 ns = .25x10 -9 s
• = 6 x 10 -2 m / .25x10 -9 s
• = 2.40000 x 10 8 m/s (raw answer)
• = 2.4 x 10 8 m/s (sig. fig. answer)
• 40. Velocity
• speed + the direction of the motion
• vector quantity
• A boat is traveling east at 15 km/hr and must pass through a current that is moving northeast at 10 km/hr . What will be the true velocity of the boat?
• 41. Acceleration
• rate of change of velocity with time
• if velocity changes there is acceleration
• includes:  v  v  direction
• formula:
 v = v f - v i units v in m/s t in s a = m/s 2 a =  v  t
• 42. Acceleration Example
• A car is traveling at 48 m/s. After 12 seconds it is traveling at 32 m/s. What is the car’s acceleration?
• a = ? 48 m/s = v i 12 s =  t 32 m/s = v f
• a =  v /  t
•    v = v f - v i = 32m/s - 48 m/s = -16 m/s
• a = -16m/s / 12 s = -1.3333333333 m/s 2
• = -1.3 m/s 2 [ -sign designates slowing down]
• 43. Application of v and a in Radiology
• KE (motion) of e- used to produce x rays
• controlling the v of e- enables the control of the photon energies
• Brems photons are produced when e - undergo a -a close to the nucleus of an atom
• 44. Newton's Laws of Motion
• 1. Inertia
• 2. Force
• 3. Recoil
• 45. Newton's First Law
• defined -- in notes
• inertia: resistance to a  in motion
• property of all matter
• mass = a measure of inertia
• 46. Inertia
• Semi-trailer truck
• large mass
• large inertia
• Bicycle
• small mass
• small inertia
• 47. Newton's 2nd Law (Force)
• Force
• anything that can  object's motion
• Fundamental forces
• Nuclear forces
• &quot;strong&quot; & &quot;weak&quot;
• Gravitational force
• Electromagnetic force
• 48. Mechanical Force
• push or pull
• vector quantity
• net force = vector sum of all forces
• push on box + friction from floor
• equilibrium -- net force = 0
Vector sum
• 49. 2nd Law (Force)
• defined -- in notes
• formula for the quantity “force”
• force = mass x acceleration
• F = m x a
Newton N a =  v  t kg m s 2
• units kg x m/s 2
• 50. Example Problem for 2nd Law
• What is the net force needed to accelerate a 5.1 kg laundry cart to 3.2 m/s 2 ?
• F =?? 5.1 kg = mass 3.2 m/s 2 = acceleration
• F = m a
• = 5.1 kg x 3.2 m/s 2
• = 16.32 kg m/s 2
• = 16 N
• 51. Example 2:
• A net force of 275 N is applied to a 110 kilogram mobile unit. What is the unit's acceleration?
• acceleration =?? 275 N = F 110 kg = mass
• F = m a
• a = F / m
• = 275[kg m/s 2 ] / 110kg
• = 2.5 m/s 2
• 52. Example 3
• An object experiences a net force of 376N. After 2 seconds the change in the object's velocity 15m/s. What is the object's mass?
• mass =?? 376 N = F 2 s =  t 15 m/s =  v
• F = m a  m = F / a
• a =  v/  t
• = 15 m / s / 2 s = 7.5 m/s 2
• m = 376 [kg m/s 2 ] / 7.5 m/s 2
• = 50.13333333333 kg = 50 kg
• 53. Weight
• adaptation of Newton's 2nd law
• weight = force caused by the pull of gravitation
• weight  mass
• gravitational force inertia of the object
• varies with gravity always constant
• unit = N [pound] unit = kg [slug]
• when g is a constant then weight proportional mass
• 54. Weight (cont.)
• formula for quantity “weight”
• modified from force formula
• F = m x a
• Wt. = m x g g earth = 9.8m/s 2
Newton N kg m s 2 units kg x m/s 2
• 55. Weight Problem
• What is the weight (on earth) of a 42 kg person?
• Wt. = ?? 42 kg = mass [9.8m/s 2 = gravity]
• Wt. = m x g
• = 42 kg x 9.8m/s 2
• = 411.6 kg m/ s 2
• = 410 N
• 56. Weight Problem #2
• What is the mass of a 2287N mobile x-ray unit?
• mass = ?? 2287N = Wt [9.8m/s 2 = gravity]
• Wt. = m x g
• m = Wt. / g
• = 2287N / 9.8m/s 2
• = 233.3673469388 kg
• = 233.4 kg
• 57. 3rd Law (Recoil)
• Defined -- in notes
• no single force in nature
• all forces act in pairs
• action vs. reaction
• formula
• F AB = -F BA
A B
• 58. Momentum (Linear)
• measures the amount of motion of an object
• tendency of an object to go in straight line when at a constant velocity
• formula
• p = m x v
• units
• = kg x m/s
• =
kg m s
• 59. Momentum vs. Mass (Inertia)
• p = m x v
• p  m
Direct proportional relationship  m =  p  m =  p
• 60. Momentum vs. Velocity
• p = m x v
• p  v
Direct proportional relationship 50 km/hr  v =  p 100 km/hr  v =  p
• 61. Momentum Problem
• What is the momentum of a 8.8 kg cart that has a speed of 1.24 m/s?
• p = ?? 8.8 kg = mass 1.24 m/s = velocity
• p = m x v
• = 8.8 kg x 1.24 m/s
• = 10.912 kg m/s
• = 11 kg m/s
• 62. Momentum Problem #2
• What is the speed of a 3.5x10 4 kg car that has a momentum of 1.4x10 5 kg m/s?
• velocity = ?? 3.5x10 4 kg = mass 1.4x10 5 kg m/s = momentum
• p = m x v
• v = p / m
• = 1.4x10 5 kg m/s / 3.5x10 4 kg
• = 4.0 x 10 0 m/s
• = 4.0 m/s
• 63. Conservation Laws
• Statements about quantities which remain the same under specified conditions.
• Most Notable Conservation Laws
• Conservation of Energy
• Conservation of Matter
• Conservation of Linear Momentum
• 64. Conservation of Linear Momentum
• momentum after a collision will equal momentum before collision
• results in a redistribution momentum among the objects
• p 1 = p 2
• m 1 v 1 = m 2 v 2
• 65. Example before collision collision occurs after collision m 1 v 1 = 1 kg m/s mv = 0 mv = 0 m 2 v 2 = 1 kg m/s
• 66. Example #2 m 1 v 1 = 5 kg m/s mv = 0 m 2 v 2 = 5 kg m/s before collision collision occurs after collision m 2 = m A + m B v 2 = v A + v B A B A B
• 67. Work
• defined -- in notes
• measures the change a force has on an object's position or motion
• If there is NO change in position or motion, NO mechanical work is done.
F d
• 68. Work (cont.)
• formula
• Work = force x distance
• W = F x d
• units = N x m
• =
kg m s 2 x m kg m 2 s 2 = Joule J =
• 69. Example
• How much mechanical work is done to lift a 12 kg mass 8.2 m off of the floor if a force of 130 N is applied?
• work = ?? 12 kg = mass 8.2 m = distance 130 N = force
• W = F x d
• = 130 N x 8.2 m
• = 1066 N m
• = 1100 J (1.1 kJ)
• 70. Example #2
• A 162 N force is used to move a 45 kg box 32 m. What is the work that is done moving the box?
• work = ?? 162 N = force 45 kg = mass 32 m = distance
• W = F x d
• = 162 N x 32 m
• = 5184 N m
• = 5200 J or 5.2 kJ
• 71. Energy
• property of matter
• enables matter to perform work
• Kinetic Energy: due to motion
• Potential Energy: due to position in a force field
• Rest Energy: due to mass
• 72. Kinetic Energy
• work done by the motion of an object
• translation, rotation, or vibration
• formula
• KE = ½ mass x velocity squared
• = ½ m v 2
• units = kg x [m/s] 2
kg m 2 s 2 = Joule J =
• 73. Example
• Find the kinetic energy of a 450 kg mobile unit moving at 6 m/s.
• kinetic energy = ?? 450 kg = mass 6 m/s = velocity
• KE = ½ m v 2
• = ½ x 450 kg x [6 m/s] 2
• = 8100 kg m 2 /s 2
• = 8000 J or 8 kJ
• 74. Potential Energy
• capacity to do work because of the object's position in a force field
• fields
• nuclear
• electromagnetic
• gravitational
• 75. Gravitational Potential Energy
• barbell with PE
• formula
• PE g = mass x gravity x height
• = m x g x h
• units
• = kg x m/s 2 x m
• =
h g m = Joule J kg m 2 s 2
• 76. Example
• How much energy does a 460 kg mobile unit possess when it is stationed on the 3rd floor of the hospital? (42m above ground)
• PE = ?? 460 kg = mass 42 m = height [9.8 m/s 2 = gravity]
• Pe g = m x g x h
• = 460 kg x 9.8 m/s 2 x 42 m
• = 189 336 kg m 2 /s 2
• = 190 000 J or 1.9x10 5 J or 190 kJ
• 77. Rest Mass Energy
• energy due to mass
• Einstein's Theory
• formula (variation of KE formula)
• E m = mass x speed of light squared
• = m c 2 [ c = 3x10 8 m/s ]
• units = kg x [m/s] 2
kg m 2 s 2 = Joule J =
• 78. Example
• What is the energy equivalent of a 2.2 kg object?
• E m = ?? 2.2 kg = mass [3x10 8 m/s = speed of light]
• E m = m c 2
• = 2.2 kg x [3x10 8 m/s ] 2
• = 1.98 x 10 17 kg m 2 /s 2
• = 2.0 x 10 17 J [trailing 0 is significant]
• 79. Conservation Of Energy (Matter)
• Energy is neither created nor destroyed but can be interchanged
• (Matter is neither created nor destroyed but can be interchanged)
• Because mass has rest energy, conservation of matter & energy can be combined
• 80. Power
• Rate at which work is done
• Faster work = more power
• Rate at which energy changes
• Large E  = more power
• 81. Power (cont.)
• formula
• power = work / time or  energy / time
• P = W / t or  E / t
• units = J / s
kg m 2 s 3 = Watt W = kg m 2 s 2 = s
• 82. Example
• How much power is used when an 80N force moves a box 15 m during a 12 s period of time?
• (hint: solve for work first)
• P = ?? 80 N = force 15 m = distance 12 s = time
• P = W / t & W = F d
• P = ( F d ) / t
• = ( 80 N x 15 m ) / 12 s
• = 100 Nm/s
• = 100 W
• 83. Heat energy
• internal kinetic energy of matter
• from the random motion of molecules or atoms
• KE & PE of molecules
• heat E in matter moves from area of higher E in object to area of lower internal E
• Unit -- Calorie (a form of the joule)
• amount of heat required to raise one gram of water one degree Celsius.
• 84. Heat Transfer
• movement of heat energy from the hotter to cooler object (or portion of object)
• 3 methods of transfer
• conduction
• convection
• 85. conduction
• primary means in solid objects
• classification of matter by heat transfer
• conductors--rapid transfer
• insulator--very slow to transfer
• 86. convection
• primary means in gasses and liquids
• convection current--continuing rise of heated g/l and sinking of cool g/l
• transfer without the use of a medium
• (i.e. no solid, liquid or gas)
• occurs in a vacuum
• term “radiation” may simply refer to heat energy and not the transfer of heat
• infra-red radiation, part of EM spectrum, is heat energy
• 89. Effects of Heat Transfer
• change in physical state of matter
• solid  liquid  gas
• melt boil
• change in temperature
• measure of the average KE of an object
• relative measure of sensible heat or cold
• 90. Temperature Scales
• Scales Boil (steam) Freeze (ice) No KE
• Fahrenheit 212° 32° -460°
• Celsius 100° 0° -273°
• Kelvin (SI) 373 273 0
• 1K = 1°C = 1.8°F
• Conversion formulae
• °F = 32 + (1.8 °C)
• °C = (°F - 32)  1.8
• K = °C + 273